Sample Solutions Manual

4 CHAPTER 1. VECTOR ANALYSIS

Chapter 1

Vector Analysis

Problem 1.1

CHAPTER 1. VECTOR ANALYSIS 3

Chapter 1

Vector Analysis

Problem 1.1

!

"

#

$

A

B

C

B+C

! "# $

|B| cos !1

! "# $

|C| cos !2

}|B| sin !1

}|C| sin !2

!1

!2

!3

(a) From the diagram, |B + C| cos !3 = |B| cos !1 + |C| cos !2.|A||B + C| cos !3 = |A||B| cos !1 + |A||C| cos !2.So: A·(B + C) = A·B + A·C. (Dot product is distributive)

Similarly: |B + C| sin !3 = |B| sin !1 + |C| sin !2. Mulitply by |A| n.|A||B + C| sin !3 n = |A||B| sin !1 n + |A||C| sin !2 n.If n is the unit vector pointing out of the page, it follows thatA!(B + C) = (A!B) + (A!C). (Cross product is distributive)

(b) For the general case, see G. E. Hay’s Vector and Tensor Analysis, Chapter 1, Section 7 (dot product) andSection 8 (cross product)

Problem 1.2

! A = B

%C

&B!C 'A!(B!C)

The triple cross-product is not in general associative. For example,suppose A = B and C is perpendicular to A, as in the diagram.Then (B!C) points out-of-the-page, and A!(B!C) points down,and has magnitude ABC. But (A!B) = 0, so (A!B)!C = 0 !=A!(B!C).

Problem 1.3

! y

%z

(x

!B

"A

!

A = +1 x + 1 y " 1 z; A =#

3; B = 1 x + 1 y + 1 z;

A·B = +1 + 1 " 1 = 1 = AB cos ! =#

3#

3 cos ! $ cos !.

! = cos!1%

13

&

% 70.5288"

Problem 1.4

The cross-product of any two vectors in the plane will give a vector perpendicular to the plane. For example,we might pick the base (A) and the left side (B):

c#2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.

(a) From the diagram, |B + C| cos θ3 = |B| cos θ1 + |C| cos θ2. Multiply by |A|.|A||B + C| cos θ3 = |A||B| cos θ1 + |A||C| cos θ2.So: A·(B + C) = A·B + A·C. (Dot product is distributive)

Similarly: |B + C| sin θ3 = |B| sin θ1 + |C| sin θ2. Mulitply by |A| n.|A||B + C| sin θ3 n = |A||B| sin θ1 n + |A||C| sin θ2 n.If n is the unit vector pointing out of the page, it follows thatA×(B + C) = (A×B) + (A×C). (Cross product is distributive)


Problem 1.2


Chapter 1

Vector Analysis

Problem 1.1

!

"

#

$

A

B

C

B+C

! "# $

|B| cos !1

! "# $

|C| cos !2

}|B| sin !1

}|C| sin !2

!1

!2

!3

(a) From the diagram, |B + C| cos !3 = |B| cos !1 + |C| cos !2. Multiply by |A|.|A||B + C| cos !3 = |A||B| cos !1 + |A||C| cos !2.So: A·(B + C) = A·B + A·C. (Dot product is distributive)



Problem 1.2

! A = B

%C

&B!C 'A!(B!C)


Problem 1.3

! y

%z

(x

!B

"A

!

A = +1 x + 1 y " 1 z; A =#

3; B = 1 x + 1 y + 1 z; B =#

3.

A·B = +1 + 1 " 1 = 1 = AB cos ! =#

3#

3 cos ! $ cos ! = 13.

! = cos!1%

13

&

% 70.5288"

Problem 1.4


A = "1 x + 2 y + 0 z; B = "1 x + 0 y + 3 z.


The triple cross-product is not in general associative. For example,suppose A = B and C is perpendicular to A, as in the diagram.Then (B×C) points out-of-the-page, and A×(B×C) points down,and has magnitude ABC. But (A×B) = 0, so (A×B)×C = 0 6=A×(B×C).

Problem 1.3


Chapter 1

Vector Analysis

Problem 1.1

!

"

#

$

A

B

C

B+C

! "# $

|B| cos !1

! "# $

|C| cos !2

}|B| sin !1

}|C| sin !2

!1

!2

!3

(a) From the diagram, |B + C| cos !3 = |B| cos !1 + |C| cos !2.|A||B + C| cos !3 = |A||B| cos !1 + |A||C| cos !2.So: A·(B + C) = A·B + A·C. (Dot product is distributive)



Problem 1.2

! A = B

%C

&B!C 'A!(B!C)


Problem 1.3

! y

%z

(x

!B

"A

!

A = +1 x + 1 y " 1 z; A =#

3; B = 1 x + 1 y + 1 z;

A·B = +1 + 1 " 1 = 1 = AB cos ! =#

3#

3 cos ! $ cos !.

! = cos!1%

13

&

% 70.5288"

Problem 1.4



A = +1 x + 1 y − 1 z; A =√

3; B = 1 x + 1 y + 1 z; B =√

3.

A·B = +1 + 1− 1 = 1 = AB cos θ =√

3√

3 cos θ ⇒ cos θ = 13 .

θ = cos−1(

13

)≈ 70.5288◦

Problem 1.4


A = −1 x + 2 y + 0 z; B = −1 x + 0 y + 3 z.

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A×B =

∣∣∣∣∣∣x y z−1 2 0−1 0 3

∣∣∣∣∣∣ = 6 x + 3 y + 2 z.

This has the right direction, but the wrong magnitude. To make a unit vector out of it, simply divide by itslength:|A×B| =

√36 + 9 + 4 = 7. n = A×B

|A×B| = 67 x + 3

7 y + 27 z .

Problem 1.5

A×(B×C) =

∣∣∣∣∣∣x y zAx Ay Az

(ByCz −BzCy) (BzCx −BxCz) (BxCy −ByCx)

∣∣∣∣∣∣= x[Ay(BxCy −ByCx)−Az(BzCx −BxCz)] + y() + z()(I’ll just check the x-component; the others go the same way)= x(AyBxCy −AyByCx −AzBzCx +AzBxCz) + y() + z().

B(A·C)−C(A·B) = [Bx(AxCx +AyCy +AzCz)− Cx(AxBx +AyBy +AzBz)] x + () y + () z= x(AyBxCy +AzBxCz −AyByCx −AzBzCx) + y() + z(). They agree.

Problem 1.6

A×(B×C)+B×(C×A)+C×(A×B) = B(A·C)−C(A·B)+C(A·B)−A(C·B)+A(B·C)−B(C·A) = 0.So: A×(B×C)− (A×B)×C = −B×(C×A) = A(B·C)−C(A·B).

If this is zero, then either A is parallel to C (including the case in which they point in opposite directions, orone is zero), or else B·C = B·A = 0, in which case B is perpendicular to A and C (including the case B = 0.)

Conclusion: A×(B×C) = (A×B)×C⇐⇒ either A is parallel to C, or B is perpendicular to A and C.

Problem 1.7

r = (4 x + 6 y + 8 z)− (2 x + 8 y + 7 z) = 2 x− 2 y + z

r =√

4 + 4 + 1 = 3

r = rr = 23 x−

23 y + 1

3 z

Problem 1.8

(a) AyBy + AzBz = (cosφAy + sinφAz)(cosφBy + sinφBz) + (− sinφAy + cosφAz)(− sinφBy + cosφBz)= cos2 φAyBy + sinφ cosφ(AyBz + AzBy) + sin2 φAzBz + sin2 φAyBy − sinφ cosφ(AyBz + AzBy) +

cos2 φAzBz

= (cos2 φ+ sin2 φ)AyBy + (sin2 φ+ cos2 φ)AzBz = AyBy +AzBz. X

(b) (Ax)2 + (Ay)2 + (Az)2 = Σ3i=1AiAi = Σ3

i=1

(Σ3

j=1RijAj

) (Σ3

k=1RikAk

)= Σj,k (ΣiRijRik)AjAk.

This equals A2x +A2

y +A2z provided Σ3

i=1RijRik ={

1 if j = k0 if j 6= k

}Moreover, if R is to preserve lengths for all vectors A, then this condition is not only sufficient but alsonecessary. For suppose A = (1, 0, 0). Then Σj,k (ΣiRijRik)AjAk = ΣiRi1Ri1, and this must equal 1 (since wewant A

2

x+A2

y+A2

z = 1). Likewise, Σ3i=1Ri2Ri2 = Σ3

i=1Ri3Ri3 = 1. To check the case j 6= k, choose A = (1, 1, 0).Then we want 2 = Σj,k (ΣiRijRik)AjAk = ΣiRi1Ri1 + ΣiRi2Ri2 + ΣiRi1Ri2 + ΣiRi2Ri1. But we alreadyknow that the first two sums are both 1; the third and fourth are equal, so ΣiRi1Ri2 = ΣiRi2Ri1 = 0, and soon for other unequal combinations of j, k. X In matrix notation: RR = 1, where R is the transpose of R.



Problem 1.9CHAPTER 1. VECTOR ANALYSIS 5

! x

"y

#z

$%

Looking down the axis:

"y

&x

'z

"z!

&y!

'x!

(

)

*

c"2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.



! x

"y

#z

$%


"y

&x

'z

"z!

&y!

'x!

(

)

*

c"2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.

A 120◦ rotation carries the z axis into the y (= z) axis, y into x (= y), and x into z (= x). So Ax = Az,Ay = Ax, Az = Ay.

R =

0 0 11 0 00 1 0

Problem 1.10

(a) No change. (Ax = Ax, Ay = Ay, Az = Az)

(b) A −→ −A, in the sense (Ax = −Ax, Ay = −Ay, Az = −Az)

(c) (A×B) −→ (−A)×(−B) = (A×B). That is, if C = A×B, C −→ C . No minus sign, in contrast tobehavior of an “ordinary” vector, as given by (b). If A and B are pseudovectors, then (A×B) −→ (A)×(B) =(A×B). So the cross-product of two pseudovectors is again a pseudovector. In the cross-product of a vectorand a pseudovector, one changes sign, the other doesn’t, and therefore the cross-product is itself a vector.Angular momentum (L = r×p) and torque (N = r×F) are pseudovectors.

(d) A·(B×C) −→ (−A)·((−B)×(−C)) = −A·(B×C). So, if a = A·(B×C), then a −→ −a; a pseudoscalarchanges sign under inversion of coordinates.Problem 1.11

(a)∇f = 2x x + 3y2 y + 4z3 z

(b)∇f = 2xy3z4 x + 3x2y2z4 y + 4x2y3z3 z

(c)∇f = ex sin y ln z x + ex cos y ln z y + ex sin y(1/z) z

Problem 1.12

(a) ∇h = 10[(2y − 6x− 18) x + (2x− 8y + 28) y]. ∇h = 0 at summit, so2y − 6x− 18 = 02x− 8y + 28 = 0 =⇒ 6x− 24y + 84 = 0

}2y − 18− 24y + 84 = 0.

22y = 66 =⇒ y = 3 =⇒ 2x− 24 + 28 = 0 =⇒ x = −2.

Top is 3 miles north, 2 miles west, of South Hadley.

(b) Putting in x = −2, y = 3:h = 10(−12− 12− 36 + 36 + 84 + 12) = 720 ft.

(c) Putting in x = 1, y = 1: ∇h = 10[(2− 6− 18) x + (2− 8 + 28) y] = 10(−22 x + 22 y) = 220(− x + y).|∇h| = 220

√2 ≈ 311 ft/mile; direction: northwest.



Problem 1.13

r = (x− x′) x + (y − y′) y + (z − z′) z; r =√

(x− x′)2 + (y − y′)2 + (z − z′)2.

(a) ∇(r 2) = ∂∂x [(x−x′)2+(y−y′)2+(z−z′)2] x+ ∂

∂y () y+ ∂∂z () z = 2(x−x′) x+2(y−y′) y+2(z−z′) z = 2 r .

(b) ∇( 1r ) = ∂∂x [(x− x′)2 + (y − y′)2 + (z − z′)2]− 1

2 x + ∂∂y ()−

12 y + ∂

∂z ()−12 z

= − 12 ()−

32 2(x− x′) x− 1

2 ()−32 2(y − y′) y − 1

2 ()−32 2(z − z′) z

= −()−32 [(x− x′) x + (y − y′) y + (z − z′) z] = −(1/r 3)r = −(1/r 2) r .

(c) ∂∂x (r n) = n r n−1 ∂ r

∂x = n r n−1( 12

1r 2 r x) = n r n−1 r x, so ∇(r n) = n r n−1 rProblem 1.14

y = +y cosφ+ z sinφ; multiply by sinφ: y sinφ = +y sinφ cosφ+ z sin2 φ.z = −y sinφ+ z cosφ; multiply by cosφ: z cosφ = −y sinφ cosφ+ z cos2 φ.Add: y sinφ+ z cosφ = z(sin2 φ+ cos2 φ) = z. Likewise, y cosφ− z sinφ = y.So ∂y

∂y = cosφ; ∂y∂z = − sinφ; ∂z

∂y = sinφ; ∂z∂z = cosφ. Therefore

(∇f)y = ∂f∂y = ∂f

∂y∂y∂y + ∂f

∂z∂z∂y = +cosφ(∇f)y + sinφ(∇f)z

(∇f)z = ∂f∂z = ∂f

∂y∂y∂z + ∂f

∂z∂z∂z = − sinφ(∇f)y + cosφ(∇f)z

}So ∇f transforms as a vector. qed

Problem 1.15

(a)∇·va = ∂∂x (x2) + ∂

∂y (3xz2) + ∂∂z (−2xz) = 2x+ 0− 2x = 0.

(b)∇·vb = ∂∂x (xy) + ∂

∂y (2yz) + ∂∂z (3xz) = y + 2z + 3x.

(c)∇·vc = ∂∂x (y2) + ∂

∂y (2xy + z2) + ∂∂z (2yz) = 0 + (2x) + (2y) = 2(x+ y)

Problem 1.16

∇·v = ∂∂x ( x

r3 ) + ∂∂y ( y

r3 ) + ∂∂z ( z

r3 ) = ∂∂x

[x(x2 + y2 + z2)−

32

]+ ∂

∂y

[y(x2 + y2 + z2)−

32

]+ ∂

∂z

[z(x2 + y2 + z2)−

32

]= ()−

32 + x(−3/2)()−

52 2x+ ()−

32 + y(−3/2)()−

52 2y + ()−

32

+ z(−3/2)()−52 2z = 3r−3− 3r−5(x2 + y2 + z2) = 3r−3− 3r−3 = 0.

This conclusion is surprising, because, from the diagram, this vector field is obviously diverging away from theorigin. How, then, can ∇·v = 0? The answer is that ∇·v = 0 everywhere except at the origin, but at theorigin our calculation is no good, since r = 0, and the expression for v blows up. In fact, ∇·v is infinite atthat one point, and zero elsewhere, as we shall see in Sect. 1.5.Problem 1.17

vy = cosφ vy + sinφ vz; vz = − sinφ vy + cosφ vz.∂vy

∂y = ∂vy

∂y cosφ+ ∂vz

∂y sinφ =(

∂vy

∂y∂y∂y + ∂vy

∂z∂z∂y

)cosφ+

(∂vz

∂y∂y∂y + ∂vz

∂z∂z∂y

)sinφ. Use result in Prob. 1.14:

=(

∂vy

∂y cosφ+ ∂vy

∂z sinφ)

cosφ+(

∂vz

∂y cosφ+ ∂vz

∂z sinφ)

sinφ.∂vz

∂z = −∂vy

∂z sinφ+ ∂vz

∂z cosφ = −(

∂vy

∂y∂y∂z + ∂vy

∂z∂z∂z

)sinφ+

(∂vz

∂y∂y∂z + ∂vz

∂z∂z∂z

)cosφ

= −(−∂vy

∂y sinφ+ ∂vy

∂z cosφ)

sinφ+(−∂vz

∂y sinφ+ ∂vz

∂z cosφ)

cosφ. So



∂vy

∂y + ∂vz

∂z = ∂vy

∂y cos2 φ+ ∂vy

∂z sinφ cosφ+ ∂vz

∂y sinφ cosφ+ ∂vz

∂z sin2 φ+ ∂vy

∂y sin2 φ− ∂vy

∂z sinφ cosφ−∂vz

∂y sinφ cosφ+ ∂vz

∂z cos2 φ

= ∂vy

∂y

(cos2 φ+ sin2 φ

)+ ∂vz

∂z

(sin2 φ+ cos2 φ

)= ∂vy

∂y + ∂vz

∂z . X

Problem 1.18

(a) ∇×va =

∣∣∣∣∣∣x y z∂∂x

∂∂y

∂∂z

x2 3xz2 −2xz

∣∣∣∣∣∣ = x(0− 6xz) + y(0 + 2z) + z(3z2 − 0) = −6xz x + 2z y + 3z2 z.

(b) ∇×vb =

∣∣∣∣∣∣x y z∂∂x

∂∂y

∂∂z

xy 2yz 3xz

∣∣∣∣∣∣ = x(0− 2y) + y(0− 3z) + z(0− x) = −2y x− 3z y − x z.

(c) ∇×vc =

∣∣∣∣∣∣x y z∂∂x

∂∂y

∂∂z

y2 (2xy + z2) 2yz

∣∣∣∣∣∣ = x(2z − 2z) + y(0− 0) + z(2y − 2y) = 0.

Problem 1.19

A x

y

z

v

v

v

vB

As we go from point A to point B (9 o’clock to 10 o’clock), xincreases, y increases, vx increases, and vy decreases, so ∂vx/∂y >0, while ∂vy/∂y < 0. On the circle, vz = 0, and there is nodependence on z, so Eq. 1.41 says

∇× v = z(∂vy

∂x− ∂vx

∂y

)points in the negative z direction (into the page), as the righthand rule would suggest. (Pick any other nearby points on thecircle and you will come to the same conclusion.) [I’m sorry, but Icannot remember who suggested this cute illustration.]

Problem 1.20

v = y x + x y; or v = yz x + xz y + xy z; or v = (3x2z − z3) x + 3 y + (x3 − 3xz2) z;or v = (sinx)(cosh y) x− (cosx)(sinh y) y; etc.

Problem 1.21

(i) ∇(fg) = ∂(fg)∂x x + ∂(fg)

∂y y + ∂(fg)∂z z =

(f ∂g

∂x + g ∂f∂x

)x +

(f ∂g

∂y + g ∂f∂y

)y +

(f ∂g

∂z + g ∂f∂z

)z

= f(

∂g∂x x + ∂g

∂y y + ∂g∂z z

)+ g

(∂f∂x x + ∂f

∂y y + ∂f∂z z

)= f(∇g) + g(∇f). qed

(iv) ∇·(A×B) = ∂∂x (AyBz −AzBy) + ∂

∂y (AzBx −AxBz) + ∂∂z (AxBy −AyBx)

= Ay∂Bz

∂x +Bz∂Ay

∂x −Az∂By

∂x −By∂Az

∂x +Az∂Bx

∂y +Bx∂Az

∂y −Ax∂Bz

∂y −Bz∂Ax

∂y

+Ax∂By

∂z +By∂Ax

∂z −Ay∂Bx

∂z −Bx∂Ay

∂z

= Bx

(∂Az

∂y −∂Ay

∂z

)+By

(∂Ax

∂z −∂Az

∂x

)+Bz

(∂Ay

∂x −∂Ax

∂y

)−Ax

(∂Bz

∂y −∂By

∂z

)−Ay

(∂Bx

∂z −∂Bz

∂x

)−Az

(∂By

∂x −∂Bx

∂y

)= B· (∇×A)−A· (∇×B). qed

(v) ∇× (fA) =(

∂(fAz)∂y − ∂(fAy)

∂z

)x +

(∂(fAx)

∂z − ∂(fAz)∂x

)y +

(∂(fAy)

∂x − ∂(fAx)∂y

)z



=(f ∂Az

∂y +Az∂f∂y − f

∂Ay

∂z −Ay∂f∂z

)x +

(f ∂Ax

∂z +Ax∂f∂z − f

∂Az

∂x −Az∂f∂x

)y

+(f

∂Ay

∂x +Ay∂f∂x − f

∂Ax

∂y −Ax∂f∂y

)z

= f[(

∂Az

∂y −∂Ay

∂z

)x +

(∂Ax

∂z −∂Az

∂x

)y +

(∂Ay

∂x −∂Ax

∂y

)z]

−[(Ay

∂f∂z −Az

∂f∂y

)x +

(Az

∂f∂x −Ax

∂f∂z

)y +

(Ax

∂f∂y −Ay

∂f∂x

)z]

= f (∇×A)−A× (∇f). qed

Problem 1.22

(a) (A·∇)B =(Ax

∂Bx

∂x +Ay∂Bx

∂y +Az∂Bx

∂z

)x +

(Ax

∂By

∂x +Ay∂By

∂y +Az∂By

∂z

)y

+(Ax

∂Bz

∂x +Ay∂Bz

∂y +Az∂Bz

∂z

)z.

(b) r = rr = x x+y y+z z√

x2+y2+z2. Let’s just do the x component.

[(r·∇)r]x = 1√(x ∂

∂x + y ∂∂y + z ∂

∂z

)x√

x2+y2+z2

= 1r

{x[

1√ + x(− 12 ) 1

(√

)3 2x]

+ yx[− 1

21

(√

)3 2y]

+ zx[− 1

21

(√

)3 2z]}

= 1r

{xr −

1r3

(x3 + xy2 + xz2

)}= 1

r

{xr −

xr3

(x2 + y2 + z2

)}= 1

r

(xr −

xr

)= 0.

Same goes for the other components. Hence: (r·∇) r = 0 .

(c) (va·∇)vb =(x2 ∂

∂x + 3xz2 ∂∂y − 2xz ∂

∂z

)(xy x + 2yz y + 3xz z)

= x2 (y x + 0 y + 3z z) + 3xz2 (x x + 2z y + 0 z)− 2xz (0 x + 2y y + 3x z)=(x2y + 3x2z2

)x +

(6xz3 − 4xyz

)y +

(3x2z − 6x2z

)z

= x2(y + 3z2

)x + 2xz

(3z2 − 2y

)y − 3x2z z

Problem 1.23

(ii) [∇(A·B)]x = ∂∂x (AxBx +AyBy +AzBz) = ∂Ax

∂x Bx +Ax∂Bx

∂x + ∂Ay

∂x By +Ay∂By

∂x + ∂Az

∂x Bz +Az∂Bz

∂x

[A×(∇×B)]x = Ay(∇×B)z −Az(∇×B)y = Ay

(∂By

∂x −∂Bx

∂y

)−Az

(∂Bx

∂z −∂Bz

∂x

)[B×(∇×A)]x = By

(∂Ay

∂x −∂Ax

∂y

)−Bz

(∂Ax

∂z −∂Az

∂x

)[(A·∇)B]x =

(Ax

∂∂x +Ay

∂∂y +Az

∂∂z

)Bx = Ax

∂Bx

∂x +Ay∂Bx

∂y +Az∂Bx

∂z

[(B·∇)A]x = Bx∂Ax

∂x +By∂Ax

∂y +Bz∂Ax

∂z

So [A×(∇×B) + B×(∇×A) + (A·∇)B + (B·∇)A]x= Ay

∂By

∂x −Ay∂Bx

∂y −Az∂Bx

∂z +Az∂Bz

∂x +By∂Ay

∂x −By∂Ax

∂y −Bz∂Ax

∂z +Bz∂Az

∂x

+Ax∂Bx

∂x +Ay∂Bx

∂y +Az∂Bx

∂z +Bx∂Ax

∂x +By∂Ax

∂y +Bz∂Ax

∂z

= Bx∂Ax

∂x +Ax∂Bx

∂x +By

(∂Ay

∂x −∂Ax

∂y/ +∂Ax

∂y/)

+Ay

(∂By

∂x −∂Bx

∂y/ +∂Bx

∂y/)

+Bz

(−∂Ax

∂z/ +∂Az

∂x + ∂Ax

∂z/)

+Az

(−∂Bx

∂z/ +∂Bz

∂x + ∂Bx

∂z/)

= [∇(A·B)]x (same for y and z)

(vi) [∇×(A×B)]x = ∂∂y (A×B)z − ∂

∂z (A×B)y = ∂∂y (AxBy −AyBx)− ∂

∂z (AzBx −AxBz)= ∂Ax

∂y By +Ax∂By

∂y −∂Ay

∂y Bx −Ay∂Bx

∂y −∂Az

∂z Bx −Az∂Bx

∂z + ∂Ax

∂z Bz +Ax∂Bz

∂z

[(B·∇)A− (A·∇)B + A(∇·B)−B(∇·A)]x= Bx

∂Ax

∂x +By∂Ax

∂y +Bz∂Ax

∂z −Ax∂Bx

∂x −Ay∂Bx

∂y −Az∂Bx

∂z +Ax

(∂Bx

∂x + ∂By

∂y + ∂Bz

∂z

)−Bx

(∂Ax

∂x + ∂Ay

∂y + ∂Az

∂z

)c©2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.


= By∂Ax

∂y +Ax

(−∂Bx

∂x/ +∂Bx

∂x/ +∂By

∂y + ∂Bz

∂z

)+Bx

(∂Ax

∂x/ −∂Ax

∂x/ −∂Ay

∂y −∂Az

∂z

)+Ay

(−∂Bx

∂y

)+Az

(−∂Bx

∂z

)+Bz

(∂Ax

∂z

)= [∇×(A×B)]x (same for y and z)

Problem 1.24

∇(f/g) = ∂∂x (f/g) x + ∂

∂y (f/g) y + ∂∂z (f/g) z

= g ∂f∂x−f ∂g

∂x

g2 x +g ∂f

∂y−f ∂g∂y

g2 y + g ∂f∂z−f ∂g

∂z

g2 z

= 1g2

[g(

∂f∂x x + ∂f

∂y y + ∂f∂z z)− f

(∂g∂x x + ∂g

∂y y + ∂g∂z z)]

= g∇f−f∇gg2 . qed

∇·(A/g) = ∂∂x (Ax/g) + ∂

∂y (Ay/g) + ∂∂z (Az/g)

= g ∂Ax∂x −Ax

∂g∂x

g2 +g

∂Ay∂y −Ay

∂g∂y

g2 + g ∂Az∂z −Az

∂g∂x

g2

= 1g2

[g(

∂Ax

∂x + ∂Ay

∂y + ∂Az

∂z

)−(Ax

∂g∂x +Ay

∂g∂y +Az

∂g∂z

)]= g∇·A−A·∇g

g2 . qed

[∇×(A/g)]x = ∂∂y (Az/g)− ∂

∂z (Ay/g)

=g ∂Az

∂y −Az∂g∂y

g2 − g∂Ay∂z −Ay

∂g∂z

g2

= 1g2

[g(

∂Az

∂y −∂Ay

∂z

)−(Az

∂g∂y −Ay

∂g∂z

)]= g(∇×A)x+(A×∇g)x

g2 (same for y and z). qed

Problem 1.25

(a) A×B =

∣∣∣∣∣∣x y zx 2y 3z3y −2x 0

∣∣∣∣∣∣ = x(6xz) + y(9zy) + z(−2x2 − 6y2)

∇·(A×B) = ∂∂x (6xz) + ∂

∂y (9zy) + ∂∂z (−2x2 − 6y2) = 6z + 9z + 0 = 15z

∇×A = x(

∂∂y (3z)− ∂

∂z (2y))

+ y(

∂∂z (x)− ∂

∂x (3z))

+ z(

∂∂x (2y)− ∂

∂y (x))

= 0; B·(∇×A) = 0

∇×B = x(

∂∂y (0)− ∂

∂z (−2x))

+ y(

∂∂z (3y)− ∂

∂x (0))

+ z(

∂∂x (−2x)− ∂

∂y (3y))

= −5 z; A·(∇×B) = −15z

∇·(A×B) ?= B·(∇×A)−A·(∇×B) = 0− (−15z) = 15z. X

(b) A·B = 3xy − 4xy = −xy ; ∇(A·B) = ∇(−xy) = x ∂∂x (−xy) + y ∂

∂y (−xy) = −y x− x y

A×(∇×B) =

∣∣∣∣∣∣x y zx 2y 3z0 0 −5

∣∣∣∣∣∣ = x(−10y) + y(5x); B×(∇×A) = 0

(A·∇)B =(x ∂

∂x + 2y ∂∂y + 3z ∂

∂z

)(3y x− 2x y) = x(6y) + y(−2x)

(B·∇)A =(3y ∂

∂x − 2x ∂∂y

)(x x + 2y y + 3z z) = x(3y) + y(−4x)

A×(∇×B) + B×(∇×A) + (A·∇)B + (B·∇)A= −10y x + 5x y + 6y x− 2x y + 3y x− 4x y = −y x− x y = ∇·(A·B). X

(c) ∇×(A×B) = x(

∂∂y (−2x2 − 6y2)− ∂

∂z (9zy))

+ y(

∂∂z (6xz)− ∂

∂x (−2x2 − 6y2))

+ z(

∂∂x (9zy)− ∂

∂y (6xz))

= x(−12y − 9y) + y(6x+ 4x) + z(0) = −21y x + 10x y

∇·A = ∂∂x (x) + ∂

∂y (2y) + ∂∂z (3z) = 1 + 2 + 3 = 6; ∇·B = ∂

∂x (3y) + ∂∂y (−2x) = 0



(B·∇)A− (A·∇)B + A(∇·B)−B(∇·A) = 3y x− 4x y − 6y x + 2x y − 18y x + 12x y = −21y x + 10x y= ∇×(A×B). X

Problem 1.26

(a) ∂2Ta

∂x2 = 2; ∂2Ta

∂y2 = ∂2Ta

∂z2 = 0 ⇒ ∇2Ta = 2.

(b) ∂2Tb

∂x2 = ∂2Tb

∂y2 = ∂2Tb

∂z2 = −Tb ⇒ ∇2Tb = −3Tb = −3 sinx sin y sin z.

(c) ∂2Tc

∂x2 = 25Tc ; ∂2Tc

∂y2 = −16Tc ; ∂2Tc

∂z2 = −9Tc ⇒ ∇2Tc = 0.

(d) ∂2vx

∂x2 = 2 ; ∂2vx

∂y2 = ∂2vx

∂z2 = 0 ⇒ ∇2vx = 2∂2vy

∂x2 = ∂2vy

∂y2 = 0 ; ∂2vy

∂z2 = 6x ⇒ ∇2vy = 6x∂2vz

∂x2 = ∂2vz

∂y2 = ∂2vz

∂z2 = 0 ⇒ ∇2vz = 0

∇2v = 2 x + 6x y.

Problem 1.27

∇·(∇×v) = ∂∂x

(∂vz

∂y −∂vy

∂z

)+ ∂

∂y

(∂vx

∂z −∂vz

∂x

)+ ∂

∂z

(∂vy

∂x −∂vx

∂y

)=(

∂2vz

∂x ∂y −∂2vz

∂y ∂x

)+(

∂2vx

∂y ∂z −∂2vx

∂z ∂y

)+(

∂2vy

∂z ∂x −∂2vy

∂x ∂z

)= 0, by equality of cross-derivatives.

From Prob. 1.18: ∇×va = −6xz x+2z y+3z2 z ⇒ ∇·(∇×va) = ∂∂x (−6xz)+ ∂

∂y (2z)+ ∂∂z (3z2) = −6z+6z = 0.

Problem 1.28

∇×(∇t) =

∣∣∣∣∣∣x y z∂∂x

∂∂y

∂∂z

∂t∂x

∂t∂y

∂t∂z

∣∣∣∣∣∣ = x(

∂2t∂y ∂z −

∂2t∂z ∂y

)+ y

(∂2t

∂z ∂x −∂2t

∂x ∂z

)+ z(

∂2t∂x ∂y −

∂2t∂y ∂x

)= 0, by equality of cross-derivatives.

In Prob. 1.11(b), ∇f = 2xy3z4 x + 3x2y2z4 y + 4x2y3z3 z, so

∇×(∇f) =

∣∣∣∣∣∣x y z∂∂x

∂∂y

∂∂z

2xy3z4 3x2y2z4 4x2y3z3

∣∣∣∣∣∣= x(3 · 4x2y2z3 − 4 · 3x2y2z3) + y(4 · 2xy3z3 − 2 · 4xy3z3) + z(2 · 3xy2z4 − 3 · 2xy2z4) = 0. X

Problem 1.29

(a) (0, 0, 0) −→ (1, 0, 0). x : 0→ 1, y = z = 0; dl = dx x;v · dl = x2 dx;∫

v · dl =∫ 1

0x2 dx = (x3/3)|10 = 1/3.

(1, 0, 0) −→ (1, 1, 0). x = 1, y : 0→ 1, z = 0; dl = dy y;v · dl = 2yz dy = 0;∫

v · dl = 0.(1, 1, 0) −→ (1, 1, 1). x = y = 1, z : 0→ 1; dl = dz z;v · dl = y2 dz = dz;

∫v · dl =

∫ 1

0dz = z|10 = 1.

Total:∫

v · dl = (1/3) + 0 + 1 = 4/3.

(b) (0, 0, 0) −→ (0, 0, 1). x = y = 0, z : 0→ 1; dl = dz z;v · dl = y2 dz = 0;∫

v · dl = 0.(0, 0, 1) −→ (0, 1, 1). x = 0, y : 0→ 1, z = 1; dl = dy y;v ·dl = 2yz dy = 2y dy;

∫v ·dl =

∫ 1

02y dy = y2|10 = 1.

(0, 1, 1) −→ (1, 1, 1). x : 0→ 1, y = z = 1; dl = dx x;v · dl = x2 dx;∫

v · dl =∫ 1

0x2 dx = (x3/3)|10 = 1/3.

Total:∫

v · dl = 0 + 1 + (1/3) = 4/3.

(c) x = y = z : 0→ 1; dx = dy = dz;v · dl = x2 dx+ 2yz dy + y2 dz = x2 dx+ 2x2 dx+ x2 dx = 4x2 dx;∫v · dl =

∫ 1

04x2 dx = (4x3/3)|10 = 4/3.

(d)∮

v · dl = (4/3)− (4/3) = 0.



Problem 1.30

x, y : 0 → 1, z = 0; da = dx dy z;v · da = y(z2 − 3) dx dy = −3y dx dy;∫

v · da = −3∫ 2

0dx∫ 2

0y dy =

−3(x|20)(y2

2 |20) = −3(2)(2) = −12. In Ex. 1.7 we got 20, for the same boundary line (the square in the

xy-plane), so the answer is no: the surface integral does not depend only on the boundary line. The total fluxfor the cube is 20 + 12 = 32.

Problem 1.31∫T dτ =

∫z2 dx dy dz. You can do the integrals in any order—here it is simplest to save z for last:∫

z2

[∫ (∫dx

)dy

]dz.

The sloping surface is x+y+z = 1, so the x integral is∫ (1−y−z)

0dx = 1−y−z. For a given z, y ranges from 0 to

1− z, so the y integral is∫ (1−z)

0(1−y− z) dy = [(1− z)y− (y2/2)]|(1−z)

0 = (1− z)2− [(1− z)2/2] = (1− z)2/2 =(1/2)− z + (z2/2). Finally, the z integral is

∫ 1

0z2( 1

2 − z + z2

2 ) dz =∫ 1

0( z2

2 − z3 + z4

2 ) dz = ( z3

6 −z4

4 + z5

10 )|10 =16 −

14 + 1

10 = 1/60.

Problem 1.32

T (b) = 1 + 4 + 2 = 7; T (a) = 0. ⇒ T (b)− T (a) = 7.

∇T = (2x+ 4y)x + (4x+ 2z3)y + (6yz2)z; ∇T ·dl = (2x+ 4y)dx+ (4x+ 2z3)dy + (6yz2)dz

(a) Segment 1: x : 0→ 1, y = z = dy = dz = 0.∫

∇T ·dl =∫ 1

0(2x) dx = x2

∣∣10

= 1.Segment 2: y : 0→ 1, x = 1, z = 0, dx = dz = 0.

∫∇T ·dl =

∫ 1

0(4) dy = 4y|10 = 4.

Segment 3: z : 0→ 1, x = y = 1, dx = dy = 0.∫

∇T ·dl =∫ 1

0(6z2) dz = 2z3

∣∣10

= 2.

∫ b

a∇T ·dl = 7. X

(b) Segment 1: z : 0→ 1, x = y = dx = dy = 0.∫

∇T ·dl =∫ 1

0(0) dz = 0.

Segment 2: y : 0→ 1, x = 0, z = 1, dx = dz = 0.∫

∇T ·dl =∫ 1

0(2) dy = 2y|10 = 2.

Segment 3: x : 0→ 1, y = z = 1, dy = dz = 0.∫

∇T ·dl =∫ 1

0(2x+ 4) dx

= (x2 + 4x)∣∣10

= 1 + 4 = 5.

∫ b

a∇T ·dl = 7. X

(c) x : 0→ 1, y = x, z = x2, dy = dx, dz = 2x dx.

∇T ·dl = (2x+ 4x)dx+ (4x+ 2x6)dx+ (6xx4)2x dx = (10x+ 14x6)dx.∫ b

a∇T ·dl =

∫ 1

0(10x+ 14x6)dx = (5x2 + 2x7)

∣∣10

= 5 + 2 = 7. X

Problem 1.33

∇·v = y + 2z + 3x∫(∇·v)dτ =

∫(y + 2z + 3x) dx dy dz =

∫∫ {∫ 2

0(y + 2z + 3x) dx

}dy dz

↪→ [(y + 2z)x+ 3

2x2]20

= 2(y + 2z) + 6

=∫ {∫ 2

0(2y + 4z + 6)dy

}dz

↪→ [y2 + (4z + 6)y

]20

= 4 + 2(4z + 6) = 8z + 16

=∫ 2

0(8z + 16)dz = (4z2 + 16z)

∣∣20

= 16 + 32 = 48.

Numbering the surfaces as in Fig. 1.29:



(i) da = dy dz x, x = 2. v·da = 2y dy dz.∫v·da =

∫∫2y dy dz = 2y2

∣∣20

= 8.(ii) da = −dy dz x, x = 0. v·da = 0.

∫v·da = 0.

(iii) da = dx dz y, y = 2. v·da = 4z dx dz.∫v·da =

∫∫4z dx dz = 16.

(iv) da = −dx dz y, y = 0. v·da = 0.∫v·da = 0.

(v) da = dx dy z, z = 2. v·da = 6x dx dy.∫v·da = 24.

(vi) da = −dx dy z, z = 0. v·da = 0.∫v·da = 0.

⇒∫v·da = 8 + 16 + 24 = 48 X

Problem 1.34

∇×v = x(0− 2y) + y(0− 3z) + z(0− x) = −2y x− 3z y − x z.da = dy dz x, if we agree that the path integral shall run counterclockwise. So(∇×v)·da = −2y dy dz.∫

(∇×v)·da =∫ {∫ 2−z

0(−2y)dy

}dz

↪→ y2∣∣2−z

0= −(2− z)2

= −∫ 2

0(4− 4z + z2)dz = −

(4z − 2z2 + z3

3

)∣∣∣20

= −(8− 8 + 8

3

)= − 8

3 -

6z

y

@@

@@

@@

y =2−z

Meanwhile, v·dl = (xy)dx+ (2yz)dy + (3zx)dz. There are three segments.

-

6z

y

@@

@@

@@

-(1)

@@I (2)

?

(3)

(1) x = z = 0; dx = dz = 0. y : 0→ 2.∫v·dl = 0.

(2) x = 0; z = 2− y; dx = 0, dz = −dy, y : 2→ 0. v·dl = 2yz dy.∫v·dl =

∫ 0

22y(2− y)dy = −

∫ 2

0(4y − 2y2)dy = −

(2y2 − 2

3y3)∣∣2

0= −

(8− 2

3 · 8)

= − 83 .

(3) x = y = 0; dx = dy = 0; z : 2→ 0. v·dl = 0.∫v·dl = 0. So

∮v·dl = − 8

3 . X

Problem 1.35

By Corollary 1,∫(∇×v)·da should equal 4

3 . ∇×v = (4z2 − 2x)x + 2z z.

(i) da = dy dz x, x = 1; y, z : 0→ 1. (∇×v)·da = (4z2 − 2)dy dz;∫(∇×v)·da =

∫ 1

0(4z2 − 2)dz

= (43z

3 − 2z)∣∣10

= 43 − 2 = − 2

3 .

(ii) da = −dx dy z, z = 0; x, y : 0→ 1. (∇×v)·da = 0;∫(∇×v)·da = 0.

(iii) da = dx dz y, y = 1; x, z : 0→ 1. (∇×v)·da = 0;∫(∇×v)·da = 0.

(iv) da = −dx dz y, y = 0; x, z : 0→ 1. (∇×v)·da = 0;∫(∇×v)·da = 0.

(v) da = dx dy z, z = 1; x, y : 0→ 1. (∇×v)·da = 2 dx dy;∫(∇×v)·da = 2.

⇒∫(∇×v)·da = − 2

3 + 2 = 43 . X



Problem 1.36

(a) Use the product rule ∇×(fA) = f(∇×A)−A× (∇f) :∫Sf(∇×A) · da =

∫S

∇×(fA) · da +∫S[A× (∇f)] · da =

∮PfA · dl +

∫S[A× (∇f)] · da. qed

(I used Stokes’ theorem in the last step.)

(b) Use the product rule ∇·(A×B) = B · (∇×A)−A · (∇×B) :∫V

B · (∇×A)dτ =∫V

∇·(A×B) dτ +∫V

A · (∇×B) dτ =∮S(A×B) · da +

∫V

A · (∇×B) dτ. qed

(I used the divergence theorem in the last step.)

Problem 1.37 r =√x2 + y2 + z2; θ = cos−1

(z√

x2+y2+z2

); φ = tan−1

(yx

).

Problem 1.38

There are many ways to do this one—probably the most illuminating way is to work it out by trigonometryfrom Fig. 1.36. The most systematic approach is to study the expression:

r = x x + y y + z z = r sin θ cosφ x + r sin θ sinφ y + r cos θ z.

If I only vary r slightly, then dr = ∂∂r (r)dr is a short vector pointing in the direction of increase in r. To make

it a unit vector, I must divide by its length. Thus:

r =∂r∂r∣∣ ∂r∂r

∣∣ ; θ =∂r∂θ∣∣ ∂r∂θ

∣∣ ; φ =∂r∂φ∣∣ ∂r∂φ

∣∣ .∂r∂r = sin θ cosφ x + sin θ sinφ y + cos θ z;

∣∣ ∂r∂r

∣∣2 = sin2 θ cos2 φ+ sin2 θ sin2 φ+ cos2 θ = 1.∂r∂θ = r cos θ cosφ x + r cos θ sinφ y − r sin θ z;

∣∣ ∂r∂θ

∣∣2 = r2 cos2 θ cos2 φ+ r2 cos2 θ sin2 φ+ r2 sin2 θ = r2.∂r∂φ = −r sin θ sinφ x + r sin θ cosφ y;

∣∣ ∂r∂φ

∣∣2 = r2 sin2 θ sin2 φ+ r2 sin2 θ cos2 φ = r2 sin2 θ.

⇒r = sin θ cosφ x + sin θ sinφ y + cos θ z.θ = cos θ cosφ x+ cos θ sinφ y− sin θ z.φ = − sinφ x + cosφ y.

Check: r·r = sin2 θ(cos2 φ+ sin2 φ) + cos2 θ = sin2 θ + cos2 θ = 1, Xθ·φ = − cos θ sinφ cosφ+ cos θ sinφ cosφ = 0, X etc.

sin θ r = sin2 θ cosφ x + sin2 θ sinφ y + sin θ cos θ z.cos θ θ = cos2 θ cosφ x + cos2 θ sinφ y − sin θ cos θ z.

Add these:(1) sin θ r + cos θ θ = + cosφ x + sinφ y;(2) φ = − sinφ x + cosφ y.

Multiply (1) by cosφ, (2) by sinφ, and subtract:

x = sin θ cosφ r + cos θ cosφ θ − sinφ φ.



Multiply (1) by sinφ, (2) by cosφ, and add:

y = sin θ sinφ r + cos θ sinφ θ + cosφ φ.

cos θ r = sin θ cos θ cosφ x + sin θ cos θ sinφ y + cos2 θ z.sin θ θ = sin θ cos θ cosφ x + sin θ cos θ sinφ y − sin2 θ z.

Subtract these:

z = cos θ r− sin θ θ.

Problem 1.39

(a) ∇·v1 = 1r2

∂∂r (r2r2) = 1

r2 4r3 = 4r∫(∇·v1)dτ =

∫(4r)(r2 sin θ dr dθ dφ) = (4)

∫ R

0r3dr

∫ π

0sin θ dθ

∫2π0 dφ = (4)

(R4

4

)(2)(2π) = 4πR4∫

v1·da =∫(r2r)·(r2 sin θ dθ dφ r) = r4

∫ π

0sin θ dθ

∫ 2π

0dφ = 4πR4 X (Note: at surface of sphere r = R.)

(b) ∇·v2 = 1r2

∂∂r

(r2 1

r2

)= 0 ⇒

∫(∇·v2)dτ = 0∫

v2·da =∫ (

1r2 r)(r2 sin θ dθ dφ r) =

∫sin θ dθ dφ = 4π.

They don’t agree! The point is that this divergence is zero except at the origin, where it blows up, so ourcalculation of

∫(∇·v2) is incorrect. The right answer is 4π.

Problem 1.40∇·v = 1

r2∂∂r (r2 r cos θ) + 1

r sin θ∂∂θ (sin θ r sin θ) + 1

r sin θ∂

∂φ (r sin θ cosφ)= 1

r2 3r2 cos θ + 1r sin θ r 2 sin θ cos θ + 1

r sin θ r sin θ(− sinφ)= 3 cos θ + 2 cos θ − sinφ = 5 cos θ − sinφ∫

(∇·v)dτ =∫(5 cos θ − sinφ) r2 sin θ dr dθ dφ =

∫ R

0r2 dr

∫ θ2

0

[∫ 2π

0(5 cos θ − sinφ) dφ

]dθ sin θ

↪→2π(5 cos θ)=(

R3

3

)(10π)

∫ π2

0sin θ cos θ dθ

↪→ sin2 θ2

∣∣∣π2

0= 1

2

= 5π3 R

3.

Two surfaces—one the hemisphere: da = R2 sin θ dθ dφ r; r = R; φ : 0→ 2π, θ : 0→ π2 .∫

v·da =∫(r cos θ)R2 sin θ dθ dφ = R3

∫ π2

0sin θ cos θ dθ

∫ 2π

0dφ = R3

(12

)(2π) = πR3.

other the flat bottom: da = (dr)(r sin θ dφ)(+θ) = r dr dφ θ (here θ = π2 ). r : 0→ R, φ : 0→ 2π.∫

v·da =∫(r sin θ)(r dr dφ) =

∫ R

0r2 dr

∫ 2π

0dφ = 2πR3

3 .

Total:∫v·da = πR3 + 2

3πR3 = 5

3πR3. X

Problem 1.41 ∇t = (cos θ + sin θ cosφ)r + (− sin θ + cos θ cosφ)θ + 1sin θ/ (− sin θ/ sinφ)φ

∇2t = ∇·(∇t)= 1

r2∂∂r

(r2(cos θ + sin θ cosφ)

)+ 1

r sin θ∂∂θ (sin θ(− sin θ + cos θ cosφ)) + 1

r sin θ∂

∂φ (− sinφ)= 1

r2 2r(cos θ + sin θ cosφ) + 1r sin θ (−2 sin θ cos θ + cos2 θ cosφ− sin2 θ cosφ)− 1

r sin θ cosφ= 1

r sin θ [2 sin θ cos θ + 2 sin2 θ cosφ− 2 sin θ cos θ + cos2 θ cosφ− sin2 θ cosφ− cosφ]= 1

r sin θ

[(sin2 θ + cos2 θ) cosφ− cosφ

]= 0.



⇒ ∇2t = 0

Check: r cos θ = z, r sin θ cosφ = x ⇒ in Cartesian coordinates t = x+ z. Obviously Laplacian is zero.

Gradient Theorem:∫ b

a∇t·dl = t(b)− t(a)

Segment 1: θ = π2 , φ = 0, r : 0→ 2. dl = dr r; ∇t·dl = (cos θ + sin θ cosφ)dr = (0 + 1)dr = dr.∫

∇t·dl =∫ 2

0dr = 2.

Segment 2: θ = π2 , r = 2, φ : 0→ π

2 . dl = r sin θ dφ φ = 2 dφ φ.

∇t·dl = (− sinφ)(2 dφ) = −2 sinφdφ.∫

∇t·dl = −∫ π

20

2 sinφdφ = 2 cosφ|π20 = −2.

Segment 3: r = 2, φ = π2 ; θ : π

2 → 0.dl = r dθ θ = 2 dθ θ; ∇t·dl = (− sin θ + cos θ cosφ)(2 dθ) = −2 sin θ dθ.∫

∇t·dl = −∫ 0

π2

2 sin θ dθ = 2 cos θ|0π2

= 2.

Total:∫ b

a∇t·dl = 2− 2 + 2 = 2 . Meanwhile, t(b)− t(a) = [2(1 + 0)]− [0( )] = 2. X

Problem 1.42 From Fig. 1.42, s = cosφ x + sinφ y; φ = − sinφ x + cosφ y; z = z

Multiply first by cosφ, second by sinφ, and subtract:s cosφ− φ sinφ = cos2 φ x + cosφ sinφ y + sin2 φ x− sinφ cosφ y = x(sin2 φ+ cos2 φ) = x.

So x = cosφ s− sinφ φ.

Multiply first by sinφ, second by cosφ, and add:s sinφ+ φ cosφ = sinφ cosφ x + sin2 φ y − sinφ cosφ x + cos2 φ y = y(sin2 φ+ cos2 φ) = y.

So y = sinφ s + cosφ φ. z = z.

Problem 1.43

(a) ∇·v = 1s

∂∂s

(s s(2 + sin2 φ)

)+ 1

s∂

∂φ (s sinφ cosφ) + ∂∂z (3z)

= 1s 2s(2 + sin2 φ) + 1

s s(cos2 φ− sin2 φ) + 3= 4 + 2 sin2 φ+ cos2 φ− sin2 φ+ 3= 4 + sin2 φ+ cos2 φ+ 3 = 8.

(b)∫(∇·v)dτ =

∫(8)s ds dφ dz = 8

∫ 2

0s ds

∫ π2

0dφ∫ 5

0dz = 8(2)

(π2

)(5) = 40π.

Meanwhile, the surface integral has five parts:top: z = 5, da = s ds dφ z; v·da = 3z s ds dφ = 15s ds dφ.

∫v·da = 15

∫ 2

0s ds

∫ π2

0dφ = 15π.

bottom: z = 0, da = −s ds dφ z; v·da = −3z s ds dφ = 0.∫v·da = 0.

back: φ = π2 , da = ds dz φ; v·da = s sinφ cosφds dz = 0.

∫v·da = 0.

left: φ = 0, da = −ds dz φ; v·da = −s sinφ cosφds dz = 0.∫v·da = 0.

front: s = 2, da = s dφ dz s; v·da = s(2 + sin2 φ)s dφ dz = 4(2 + sin2 φ)dφ dz.∫v·da = 4

∫ π2

0(2 + sin2 φ)dφ

∫ 5

0dz = (4)(π + π

4 )(5) = 25π.

So∮v·da = 15π + 25π = 40π. X

(c) ∇×v =(

1s

∂∂φ (3z)− ∂

∂z (s sinφ cosφ))s +

(∂∂z

(s(2 + sin2 φ)

)− ∂

∂s (3z))φ

+ 1s

(∂∂s (s2 sinφ cosφ)− ∂

∂φ

(s(2 + sin2 φ)

))z

= 1s (2s sinφ cosφ− s 2 sinφ cosφ) z = 0.



Problem 1.44

(a) 3(32)− 2(3)− 1 = 27− 6− 1 = 20.

(b) cosπ = -1.

(c) zero.

(d) ln(−2 + 3) = ln 1 = zero.

Problem 1.45

(a)∫ 2

−2(2x+ 3) 1

3δ(x) dx = 13 (0 + 3) = 1.

(b) By Eq. 1.94, δ(1− x) = δ(x− 1), so 1 + 3 + 2 = 6.

(c)∫ 1

−19x2 1

3 δ(x+ 13 ) dx = 9

(− 1

3

)2 13 = 1

3 .

(d) 1 (if a > b), 0 (if a < b).

Problem 1.46

(a)∫∞−∞ f(x)

[x d

dxδ(x)]dx = x f(x)δ(x)|∞−∞ −

∫∞−∞

ddx (x f(x)) δ(x) dx.

The first term is zero, since δ(x) = 0 at ±∞; ddx (x f(x)) = x df

dx + dxdxf = x df

dx + f.

So the integral is −∫∞−∞

(x df

dx + f)δ(x) dx = 0− f(0) = −f(0) = −

∫∞−∞ f(x)δ(x) dx.

So, x ddxδ(x) = −δ(x). qed

(b)∫∞−∞ f(x) dθ

dxdx = f(x)θ(x)|∞−∞ −∫∞−∞

dfdxθ(x)dx = f(∞)−

∫∞0

dfdxdx = f(∞)− (f(∞)− f(0))

= f(0) =∫∞−∞ f(x)δ(x) dx. So dθ

dx = δ(x). qed

Problem 1.47

(a) ρ(r) = qδ3(r− r′). Check:∫ρ(r)dτ = q

∫δ3(r− r′) dτ = q. X

(b) ρ(r) = qδ3(r− a)− qδ3(r).

(c) Evidently ρ(r) = Aδ(r −R). To determine the constant A, we require

Q =∫ρ dτ =

∫Aδ(r −R)4πr2 dr = A 4πR2. So A = Q

4πR2 . ρ(r) = Q4πR2 δ(r −R).

Problem 1.48

(a) a2 + a·a + a2 = 3a2.

(b)∫(r− b)2 1

53 δ3(r) dτ = 1

125b2 = 1

125 (42 + 32) = 15 .

(c) c2 = 25 + 9 + 4 = 38 > 36 = 62, so c is outside V, so the integral is zero.

(d) (e− (2 x + 2 y + 2 z))2 = (1 x + 0 y + (−1) z)2 = 1 + 1 = 2 < (1.5)2 = 2.25, so e is inside V,and hence the integral is e·(d− e) = (3, 2, 1)·(−2, 0, 2) = −6 + 0 + 2 = -4.

Problem 1.49First method: use Eq. 1.99 to write J =

∫e−r

(4πδ3(r)

)dτ = 4πe−0 = 4π.

Second method: integrating by parts (use Eq. 1.59).



J = −∫V

rr2·∇(e−r) dτ +

∮S

e−r rr2· da. But ∇

(e−r)

=(∂

∂re−r

)r = −e−r r.

=∫

1r2e−r4πr2 dr +

∫e−r r

r2· r2 sin θ dθ dφ r = 4π

R∫0

e−r dr + e−R

∫sin θ dθ dφ

= 4π(−e−r

)∣∣R0

+ 4πe−R = 4π(−e−R + e−0

)+ 4πe−R = 4π.X

(Here R =∞, so e−R = 0.

)

Problem 1.50 (a) ∇·F1 = ∂∂x (0) + ∂

∂y (0) + ∂∂z

(x2)

= 0 ; ∇·F2 = ∂x∂x + ∂y

∂y + ∂z∂z = 1 + 1 + 1 = 3

∇×F1 =

∣∣∣∣∣∣x y z∂∂x

∂∂y

∂∂z

0 0 x2

∣∣∣∣∣∣ = −y∂

∂x

(x2)

= −2xy ; ∇×F2 =

∣∣∣∣∣∣x y z∂∂x

∂∂y

∂∂z

x y z

∣∣∣∣∣∣ = 0

F2 is a gradient; F1 is a curl U2 = 12

(x3 + y2 + z2

)would do (F2 = ∇U2).

For A1, we want(

∂Ay

∂z −∂Az

∂y

)=(

∂Ax

∂z −∂Az

∂x

)= 0; ∂Ay

∂x −∂Ax

∂y = x2. Ay = x3

3 , Ax = Az = 0 would do it.

A1 = 13x

2 y (F1 = ∇×A1) . (But these are not unique.)

(b) ∇·F3 = ∂∂x (yz) + ∂

∂y (xz) + ∂∂z (xy) = 0; ∇×F3 =

∣∣∣∣∣∣x y z∂∂x

∂∂y

∂∂z

yz xz xy

∣∣∣∣∣∣ = x (x − x) + y (y − y) + z (z − z) = 0.

So F3 can be written as the gradient of a scalar (F3 = ∇U3) and as the curl of a vector (F3 = ∇×A3). Infact, U3 = xyz does the job. For the vector potential, we have

∂Az

∂y −∂Ay

∂z = yz, which suggests Az = 14y

2z + f(x, z); Ay = − 14yz

2 + g(x, y)∂Ax

∂z −∂Az

∂x = xz, suggesting Ax = 14z

2x+ h(x, y); Az = − 14zx

2 + j(y, z)∂Ay

∂x −∂Ax

∂y = xy, so Ay = 14x

2y + k(y, z); Ax = − 14xy

2 + l(x, z)

Putting this all together: A3 = 1

4

{x(z2 − y2

)x + y

(x2 − z2

)y + z

(y2 − x2

)z}

(again, not unique).

Problem 1.51(d) ⇒ (a): ∇×F = ∇×(−∇U) = 0 (Eq. 1.44 – curl of gradient is always zero).(a) ⇒ (c):

∮F · dl =

∫(∇×F) · da = 0 (Eq. 1.57–Stokes’ theorem).

(c) ⇒ (b):∫ b

a IF · dl−

∫ b

a IIF · dl =

∫ b

a IF · dl +

∫ a

b IIF · dl =

∮F · dl = 0, so∫ b

a I

F · dl =∫ b

a II

F · dl.

(b) ⇒ (c): same as (c) ⇒ (b), only in reverse; (c) ⇒ (a): same as (a)⇒ (c).

Problem 1.52(d) ⇒ (a): ∇·F = ∇·(∇×W) = 0 (Eq 1.46—divergence of curl is always zero).(a) ⇒ (c):

∮F · da =

∫(∇·F) dτ = 0 (Eq. 1.56—divergence theorem).



(c) ⇒ (b):∫

IF · da−

∫II

F · da =∮

F · da = 0, so∫I

F · da =∫

II

F · da.

(Note: sign change because for∮

F · da, da is outward, whereas for surface II it is inward.)(b) ⇒ (c): same as (c) ⇒ (b), in reverse; (c)⇒ (a): same as (a)⇒ (c) .

Problem 1.53In Prob. 1.15 we found that ∇·va = 0; in Prob. 1.18 we found that ∇×vc = 0. So

vc can be written as the gradient of a scalar; va can be written as the curl of a vector.

(a) To find t:

(1) ∂t∂x = y2 ⇒ t = y2x+ f(y, z)

(2) ∂t∂y =

(2xy + z2

)(3) ∂t

∂z = 2yz

From (1) & (3) we get ∂f∂z = 2yz ⇒ f = yz2 + g(y) ⇒ t = y2x + yz2 + g(y), so ∂t

∂y = 2xy + z2 + ∂g∂y =

2xy + z2 (from (2)) ⇒ ∂g∂y = 0. We may as well pick g = 0; then t = xy2 + yz2.

(b) To find W: ∂Wz

∂y −∂Wy

∂z = x2; ∂Wx

∂z −∂Wz

∂x = 3z2x; ∂Wy

∂x −∂Wx

∂y = −2xz.

Pick Wx = 0; then

∂Wz

∂x= −3xz2 ⇒Wz = −3

2x2z2 + f(y, z)

∂Wy

∂x= −2xz ⇒Wy = −x2z + g(y, z).

∂Wz

∂y −∂Wy

∂z = ∂f∂y + x2 − ∂g

∂z = x2 ⇒ ∂f∂y −

∂g∂z = 0. May as well pick f = g = 0.

W = −x2z y − 32x

2z2 z.

Check: ∇×W =

∣∣∣∣∣∣x y z∂∂x

∂∂y

∂∂z

0 −x2z − 32x

2z2

∣∣∣∣∣∣ = x(x2)

+ y(3xz2

)+ z (−2xz).X

You can add any gradient (∇t) to W without changing its curl, so this answer is far from unique. Someother solutions:

W = xz3 x− x2z y;

W =(2xyz + xz3

)x + x2y z;

W = xyz x− 12x

2z y + 12x

2(y − 3z2

)z.



Problem 1.54

∇·v =1r2

∂

∂r

(r2 r2 cos θ

)+

1r sin θ

∂

∂θ

(sin θ r2 cosφ

)+

1r sin θ

∂

∂φ

(−r2 cos θ sinφ

)=

1r2

4r3 cos θ +1

r sin θcos θ r2 cosφ+

1r sin θ

(−r2 cos θ cosφ

)=r cos θsin θ

[4 sin θ + cosφ− cosφ] = 4r cos θ.

∫(∇·v) dτ =

∫(4r cos θ)r2 sin θ dr dθ dφ = 4

R∫0

r3 dr

π/2∫0

cos θ sin θ dθ

π/2∫0

dφ

=(R4)(1

2

)(π2

)= πR4

4 .

Surface consists of four parts:(1) Curved: da = R2 sin θ dθ dφ r; r = R. v · da =

(R2 cos θ

) (R2 sin θ dθ dφ

).

∫v · da = R4

π/2∫0

cos θ sin θ dθ

π/2∫0

dφ = R4

(12

)(π2

)=πR4

4.

(2) Left: da = −r dr dθ φ; φ = 0. v · da =(r2 cos θ sinφ

)(r dr dθ) = 0.

∫v · da = 0.

(3) Back: da = r dr dθ φ; φ = π/2. v · da =(−r2 cos θ sinφ

)(r dr dθ) = −r3 cos θ dr dθ.

∫v · da =

R∫0

r3 dr

π/2∫0

cos θ dθ = −(

14R4

)(+1) = −1

4R4.

(4) Bottom: da = r sin θ dr dφ θ; θ = π/2. v · da =(r2 cosφ

)(r dr dφ) .

∫v · da =

R∫0

r3 dr

π/2∫0

cosφdφ =14R4.

Total:∮

v · da = πR4/4 + 0− 14R

4 + 14R

4 = πR4

4 . X

Problem 1.55

∇×v =

∣∣∣∣∣∣x y z∂∂x

∂∂y

∂∂z

ay bx 0

∣∣∣∣∣∣ = z (b− a). So∫

(∇×v) · da = (b− a)πR2.

v · dl = (ay x + bx y) · (dx x + dy y + dz z) = ay dx+ bx dy; x2 + y2 = R2 ⇒ 2x dx+ 2y dy = 0,so dy = −(x/y) dx. So v · dl = ay dx+ bx(−x/y) dx = 1

y

(ay2 − bx2

)dx.



For the “upper” semicircle, y =√R2 − x2, so v · dl =

a(R2−x2)−bx2

√R2−x2 dx.

∫v · dl =

−R∫R

aR2 − (a+ b)x2

√R2 − x2

dx ={aR2 sin−1

( xR

)− (a+ b)

[−x

2

√R2 − x2 +

R2

2sin−1

( xR

)]}∣∣∣∣−R

+R

=12R2(a− b) sin−1(x/R)

∣∣∣∣−R

+R

=12R2(a− b)

(sin−1(−1)− sin−1(+1)

)=

12R2(a− b)

(−π

2− π

2

)=

12πR2(b− a).

And the same for the lower semicircle (y changes sign, but the limits on the integral are reversed) so∮v · dl = πR2(b− a). X

Problem 1.56(1) x = z = 0; dx = dz = 0; y : 0→ 1. v · dl = (yz2) dy = 0;

∫v · dl = 0.

(2) x = 0; z = 2−2y; dz = −2 dy; y : 1→ 0. v ·dl = (yz2) dy+(3y+z) dz = y(2−2y)2 dy−(3y+2−2y)2 dy;

∫v · dl = 2

0∫1

(2y3 − 4y2 + y − 2) dy = 2(y4

2− 4y3

3+y2

2− 2y

)∣∣∣∣01

=143.

(3) x = y = 0; dx = dy = 0; z : 2→ 0. v · dl = (3y + z) dz = z dz;

∫v · dl =

0∫2

z dz =z2

2

∣∣∣02

= −2.

Total:∮

v · dl = 0 + 143 − 2 = 8

3 .

Meanwhile, Stokes’ thereom says∮

v · dl =∫

(∇×v) · da. Here da = dy dz x, so all we need is(∇×v)x = ∂

∂y (3y + z)− ∂∂z (yz2) = 3− 2yz. Therefore

∫(∇×v) · da =

∫ ∫(3− 2yz) dy dz =

∫ 1

0

[∫ 2−2y

0

(3− 2yz) dz]dy

=∫ 1

0

[3(2− 2y)− 2y 1

2 (2− 2y)2]dy =

∫ 1

0

(−4y3 + 8y2 − 10y + 6) dy

=(−y4 + 8

3y3 − 5y2 + 6y

)∣∣∣10

= −1 + 83 − 5 + 6 = 8

3 . X

Problem 1.57Start at the origin.

(1) θ = π2 , φ = 0; r : 0→ 1. v · dl =

(r cos2 θ

)(dr) = 0.

∫v · dl = 0.

(2) r = 1, θ = π2 ; φ : 0→ π/2. v · dl = (3r)(r sin θ dφ) = 3 dφ.

∫v · dl = 3

π/2∫0

dφ = 3π2 .



(3) φ = π2 ; r sin θ = y = 1, so r = 1

sin θ , dr = −1sin2 θ

cos θ dθ, θ : π2 → θ0 ≡ tan−1(1/2).

v · dl =(r cos2 θ

)(dr)− (r cos θ sin θ)(r dθ) =

cos2 θsin θ

(− cos θ

sin2 θ

)dθ − cos θ sin θ

sin2 θdθ

= −(

cos3 θsin3 θ

+cos θsin θ

)dθ = −cos θ

sin θ

(cos2 θ + sin2 θ

sin2 θ

)dθ = − cos θ

sin3 θdθ.

Therefore ∫v · dl = −

θ0∫π/2

cos θsin3 θ

dθ =1

2 sin2 θ

∣∣∣∣θ0

π/2

=1

2 · (1/5)− 1

2 · (1)=

52− 1

2= 2.

(4) θ = θ0, φ = π2 ; r :

√5→ 0. v · dl =

(r cos2 θ

)(dr) = 4

5r dr.

∫v · dl =

45

0∫√

5

r dr =45r2

2

∣∣∣∣0√5

= −45· 52

= −2.

Total: ∮v · dl = 0 +

3π2

+ 2− 2 = 3π2 .

Stokes’ theorem says this should equal∫

(∇×v) · da

∇×v =1

r sin θ

[∂

∂θ(sin θ 3r)− ∂

∂φ(−r sin θ cos θ)

]r +

1r

[1

sin θ∂

∂φ

(r cos2 θ

)− ∂

∂r(r3r)

]θ

+1r

[∂

∂r(−rr cos θ sin θ)− ∂

∂θ

(r cos2 θ

)]φ

=1

r sin θ[3r cos θ] r +

1r[−6r] θ +

1r[−2r cos θ sin θ + 2r cos θ sin θ] φ

= 3 cot θ r− 6 θ.

(1) Back face: da = −r dr dθ φ; (∇×v) · da = 0.∫

(∇×v) · da = 0.

(2) Bottom: da = −r sin θ dr dφ θ; (∇×v) · da = 6r sin θ dr dφ. θ = π2 , so (∇×v) · da = 6r dr dφ

∫(∇×v) · da =

1∫0

6r dr

π/2∫0

dφ = 6 · 12· π

2=

3π2. X

Problem 1.58v · dl = y dz.

(1) Left side: z = a− x; dz = −dx; y = 0. Therefore∫

v · dl = 0.

(2) Bottom: dz = 0. Therefore∫

v · dl = 0.



(3) Back: z = a− 12y; dz = −1/2 dy; y : 2a→ 0.

∫v · dl =

0∫2a

y(− 1

2 dy)

= − 12

y2

2

∣∣∣02a

= 4a2

4 = a2.

Meanwhile, ∇×v = x, so∫

(∇×v) · da is the projection of this surface on the x y plane = 12 · a · 2a = a2. X

Problem 1.59

∇·v =1r2

∂

∂r

(r2r2 sin θ

)+

1r sin θ

∂

∂θ

(sin θ 4r2 cos θ

)+

1r sin θ

∂

∂φ

(r2 tan θ

)=

1r2

4r3 sin θ +1

r sin θ4r2

(cos2 θ − sin2 θ

)=

4rsin θ

(sin2 θ + cos2 θ − sin2 θ

)= 4r

cos2 θsin θ

.

∫(∇·v) dτ =

∫ (4r

cos2 θsin θ

)(r2 sin θ dr dθ dφ

)=

R∫0

4r3 dr

π/6∫0

cos2 θ dθ

2π∫0

dφ =(R4)(2π)

[θ

2+

sin 2θ4

]∣∣∣∣π/6

0

= 2πR4

(π

12+

sin 60◦

4

)=πR4

6

(π + 3

√3

2

)= πR4

12

(2π + 3

√3).

Surface coinsists of two parts:

(1) The ice cream: r = R; φ : 0→ 2π; θ : 0→ π/6; da = R2 sin θ dθ dφ r; v·da =(R2 sin θ

) (R2 sin θ dθ dφ

)=

R4 sin2 θ dθ dφ.

∫v·da = R4

π/6∫0

sin2 θ dθ

2π∫0

dφ =(R4)(2π)

[12θ − 1

4sin 2θ

]π/6

0

= 2πR4

(π

12− 1

4sin 60◦

)=πR4

6

(π − 3

√3

2

)

(2) The cone: θ = π6 ; φ : 0→ 2π; r : 0→ R; da = r sin θ dφ dr θ =

√3

2 r dr dφ θ; v · da =√

3 r3 dr dφ

∫v · da =

√3

R∫0

r3 dr

2π∫0

dφ =√

3 · R4

4· 2π =

√3

2πR4.

Therefore∫

v · da = πR4

2

(π3 −

√3

2 +√

3)

= πR4

12

(2π + 3

√3). X.

Problem 1.60(a) Corollary 2 says

∮(∇T )·dl = 0. Stokes’ theorem says

∮(∇T )·dl =

∫[∇×(∇T )]·da. So

∫[∇×(∇T )]·da = 0,

and since this is true for any surface, the integrand must vanish: ∇×(∇T ) = 0, confirming Eq. 1.44.

(b) Corollary 2 says∮

(∇×v)·da = 0.Divergence theorem says∮

(∇×v)·da =∫

∇·(∇×v) dτ. So∫

∇·(∇×v) dτ= 0, and since this is true for any volume, the integrand must vanish: ∇(∇×v) = 0, confirming Eq. 1.46.

Problem 1.61(a) Divergence theorem:

∮v · da =

∫(∇·v) dτ. Let v = cT , where c is a constant vector. Using product

rule #5 in front cover: ∇·v = ∇·(cT ) = T (∇·c)+ c · (∇T ). But c is constant so ∇·c = 0. Therefore we have:∫c · (∇T ) dτ =

∫Tc · da. Since c is constant, take it outside the integrals: c ·

∫∇T dτ = c ·

∫T da. But c



is any constant vector—in particular, it could be be x, or y, or z—so each component of the integral on leftequals corresponding component on the right, and hence∫

∇T dτ =∫T da. qed

(b) Let v → (v × c) in divergence theorem. Then∫

∇·(v × c)dτ =∫

(v × c) · da. Product rule #6 ⇒∇·(v × c) = c · (∇×v) − v · (∇×c) = c · (∇×v). (Note: ∇×c = 0, since c is constant.) Meanwhile vectorindentity (1) says da · (v × c) = c · (da× v) = −c · (v × da). Thus

∫c · (∇×v) dτ = −

∫c · (v × da). Take c

outside, and again let c be x, y, z then:∫(∇×v) dτ = −

∫v × da. qed

(c) Let v = T∇U in divergence theorem:∫

∇·(T∇U) dτ =∫T∇U ·da. Product rule #(5)⇒∇·(T∇U) =

T∇·(∇U) + (∇U) · (∇T ) = T∇2U + (∇U) · (∇T ). Therefore∫ (T∇2U + (∇U) · (∇T )

)dτ =

∫(T∇U) · da. qed

(d) Rewrite (c) with T ↔ U :∫ (U∇2T + (∇T ) · (∇U)

)dτ =

∫(U∇T ) ·da. Subtract this from (c), noting

that the (∇U) · (∇T ) terms cancel:∫ (T∇2U − U∇2T

)dτ =

∫(T∇U − U∇T ) · da. qed

(e) Stokes’ theorem:∫

(∇×v) · da =∮

v · dl. Let v = cT . By Product Rule #(7): ∇×(cT ) = T (∇×c)−c× (∇T ) = −c× (∇T ) (since c is constant). Therefore, −

∫(c× (∇T )) · da =

∮Tc · dl. Use vector indentity

#1 to rewrite the first term (c× (∇T )) · da = c · (∇T × da). So −∫

c · (∇T × da) =∮

c ·T dl. Pull c outside,and let c→ x, y, and z to prove: ∫

∇T × da = −∮T dl. qed

Problem 1.62(a) da = R2 sin θ dθ dφ r. Let the surface be the northern hemisphere. The x and y components clearly integrateto zero, and the z component of r is cos θ, so

a =∫R2 sin θ cos θ dθ dφ z = 2πR2 z

∫ π/2

0

sin θ cos θ dθ = 2πR2 zsin2 θ

2

∣∣∣π/2

0= πR2 z.

(b) Let T = 1 in Prob. 1.61(a). Then ∇T = 0, so∮da = 0. qed

(c) This follows from (b). For suppose a1 6= a2; then if you put them together to make a closed surface,∮da = a1 − a2 6= 0.

(d) For one such triangle, da = 12 (r × dl) (since r × dl is the area of the parallelogram, and the direction is

perpendicular to the surface), so for the entire conical surface, a = 12

∮r× dl.

(e) Let T = c · r, and use product rule #4: ∇T = ∇(c · r) = c × (∇×r) + (c ·∇)r. But ∇×r = 0, and(c ·∇)r = (cx ∂

∂x + cy∂∂y + cz

∂∂z )(x x + y y + z z) = cx x + cy y + cz z = c. So Prob. 1.61(e) says∮

T dl =∮

(c · r) dl = −∫

(∇T )× da = −∫

c× da = −c×∫da = −c× a = a× c. qed



Problem 1.63

(1)

∇·v =1r2

∂

∂r

(r2 · 1

r

)=

1r2

∂

∂r(r) = 1

r2 .

For a sphere of radius R:∫v · da =

∫ (1R r)·(R2 sin θ dθ dφ r

)= R

∫sin θ dθ dφ = 4πR.∫

(∇·v) dτ =∫ (

1r2

) (r2 sin θ dr dθ dφ

)=

(R∫0

dr

)(∫sin θ dθ dφ

)= 4πR.

So divergencetheorem checks.

Evidently there is no delta function at the origin.

∇× (rn r) =1r2

∂

∂r

(r2rn

)=

1r2

∂

∂r

(rn+2

)=

1r2

(n+ 2)rn+1 = (n+ 2)rn−1

(except for n = −2, for which we already know (Eq. 1.99) that the divergence is 4πδ3(r)).

(2) Geometrically, it should be zero. Likewise, the curl in the spherical coordinates obviously gives zero.To be certain there is no lurking delta function here, we integrate over a sphere of radius R, usingProb. 1.61(b): If ∇×(rn r) = 0, then

∫(∇×v) dτ = 0 ?= −

∮v × da. But v = rn r and da =

R2 sin θ dθ dφ r are both in the r directions, so v × da = 0. X

Problem 1.64(a) Since the argument is not a function of angle, Eq. 1.73 says

D = − 14π

1r2

d

dr

[r2(−1

2

)2r

(r2 + ε2)3/2

]=

14πr2

d

dr

[r3

(r2 + ε2)3/2

]=

14πr2

[3r2

(r2 + ε2)3/2− 3

2r3 2r

(r2 + ε2)5/3

]=

14πr2

3r2

(r2 + ε2)5/2

(r2 + ε2 − r2

)=

3ε2

4π(r2 + ε2)5/2. X

(b) Setting r → 0:

D(0, ε) =3ε2

4πε5=

34πε3

,

which goes to infinity as ε→ 0. X(c) From (a) it is clear that D(r, 0) = 0 for r 6= 0. X(d) ∫

D(r, ε) 4πr2 dr = 3ε2∫ ∞

0

r2

(r2 + ε2)5/2dr = 3ε2

(1

3ε2

)= 1. X

(I looked up the integral.) Note that (b), (c), and (d) are the defining conditions for δ3(r).


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